Question

How do you solve \[25{e^x} = 1000\]?

Answer 1

Hint: In this question we have to find value of \[x\], in doing so we have to use the concept of logarithms, first we will simplify the equation by taking all constants to one side and all \[x\] terms to other side, and now apply logarithms on both sides and applying logarithmic identity i.e.,\[\ln {a^x} = x\ln a\] and simplifying the expression we will get the required value for \[x\]. And remember that the value of \[\ln e\] is equal to 1.

Complete step-by-step solution:
A logarithm is an exponent which indicates to what power a base must be raised to produce a given number.
 \[y = {b^x}\] exponential form,
\[x = {\log _b}y\] logarithmic function, where \[x\] is the logarithm of \[y\] to the base \[b\], and\[{\log _b}y\] is the power to which we have to raise \[b\] to get \[y\], we are expressing\[x\] in terms of \[y\].
Given expression is \[25{e^x} = 1000\],
Now dividing both sides with 25 we get,
\[ \Rightarrow \dfrac{{25{e^x}}}{{25}} = \dfrac{{1000}}{{25}}\],
Now eliminating the like terms we get,
\[ \Rightarrow {e^x} = 40\],
Now applying logarithms on both sides we get,
\[ \Rightarrow \ln {e^x} = \ln 40\],
Now applying logarithmic identity \[\ln {a^x} = x\ln a\], we get,
\[ \Rightarrow x\ln e = \ln 40\],
We know that\[\ln e = 1\], now substituting the value in the above expression we get,
\[ \Rightarrow x = \ln 40\],
So, the value of \[x\] is equal to \[\ln 40\].

\[\therefore \]The value of \[x\] in the given expression \[25{e^x} = 1000\] is equal to \[\ln 40\].

Note: A logarithm is a mathematical operation that determines how many times a certain number, called the base, is multiplied by itself to reach another number, in these types of questions, we use logarithmic properties and formulas, and some of useful formulas are:
 \[{\log _a}xy = {\log _a}x + {\log _a}y\],
\[\log x - \log y = \log \left( {\dfrac{x}{y}} \right)\]
\[{\log _a}{x^n} = n{\log _a}x\],
\[{\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}\],
\[{\log _{\dfrac{1}{a}}}b = - {\log _a}b\],
\[{\log _a}a = 1\],
\[{\log _{{a^x}}}b = \dfrac{1}{x}{\log _a}b\]